Is x an integer?

Question

Is x an integer?

(1) x^2 is an integer
(2) 2x is an integer

IanStewart · Answer

This is actually a much more difficult question (at least if you want a proper mathematical proof) then it might appear to be. It's probably easier to follow the solution below if you first look at some numerical examples. If Statement 2 is true, but the answer to the question is 'no' -- that is, if x is  not  an integer -- then 2x must be odd. So in those cases, 2x might equal something like 3 or 5 or 11, and x would equal something like 3/2 or 5/2 or 11/2. But notice in each case, x^2 will never be an integer -- if you square each of these, you'll get 2.25, or 6.25, or 30.25. In fact x^2 will always end in '.25' if you write it as a decimal.

So we can show that the two statements are sufficient if we start out by assuming we can get a 'no' answer, and then prove that no value of x can exist where both statements are true.

Look first at Statement 2. If the answer is 'no', so x is not an integer, then 2x is odd. Then x itself will be equal to some integer plus 1/2 (as a decimal, it will end in ".5"), because if 2x is odd, the remainder is 1 when we divide 2x by 2.

So if the answer to the question is 'no', then x must be equal to something like k + 1/2, where k is an integer. But if we also assume Statement 1 is true, we know x^2 is also an integer, so if the answer to the question is 'no', (k + 1/2)^2 would need to be an integer. But (k + 1/2)^2 = k^2 + k + (1/4), and since k is an integer, this is exactly 1/4 more than some integer (its decimal would end in ".25"), so it is impossible for this to be an integer.

That is, if you assume the answer to the question is 'no' and use both statements, you find it's impossible that both statements are true. That means the answer must be 'yes', so C is the answer (since neither statement is sufficient alone).

chetan2u · Answer

1) statement 1 tells us that x^2 is an integer.. so x can be an integer or x can be \sqrt{}  of some integer.... insuff
2) stat 2 tells us that 2x is an integer... here x can be an integer or x can be a fraction with denominator 2.... insuff

combined... if x is a fraction, then x^2 will be a fraction with 4 as denominator and hence not an integer ..hence will not satisfy our statement 1..
only scenario for both stat 1 and 2 to be true is when x is an integer.. so sufficient
ans C

Bunuel · Answer

Is x an integer?

(1) x^2 is an integer --> x is either an integer (..., -1, 0, 1, 2, ...) or  \sqrt{integer} , for example  \sqrt{2} . Not sufficient.  Notice here that x cannot be some reduced fraction like 1/3 or 13/5, because in this case x^2 won't be an integer.

(2) 2x is an integer --> x is either an integer or integers/2, for example: 1/2, 3/2, ... Not sufficient.  Notice here that x cannot be some irrational number like  \sqrt{2}  or  \sqrt{integer} , because in this case 2x won't be an integer.

(1)+(2) Since both x^2 and 2x are integers then, as discussed above, x must be an integer. Sufficient.

Answer: C.

Bunuel · Answer

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camiu · Answer

Thanks everybody for your comments, now i get it!

OptimusPrepJanielle · Answer

Is x an integer?

(1) x^2 is an integer
 x^2 = I, where I is some integer. Take the square root of both sides to see that x=sqr root I. The square root of I can be may or may not produce an integer (square root of 4 or square root of 5 for example). Not sufficient. 
(2) 2x is an integer  2x = I => x= (I/2). If that integer is 8, x is an integer. If that integer is 5, x is not an integer. Not sufficient.

Statements 1 and 2, however are sufficient. 
since 2x=I, you can substitute it into x^2 = I => x^2 = 2x. Dividing both sides by x yields an integer for x. Sufficient - yes.

C

IanStewart · Answer

You're assuming that x^2 and 2x are equal to the  same  integer (that's the only way to get the equation x^2 = 2x), but that is not something we are told. Notice that in your solution, you actually found a value for x (you found that x=2) which cannot be right, because it's clear x can be any integer at all using both statements.

OptimusPrepJanielle · Answer

Good point. I had actually considered that. Still, if you know that x = some integer/2, and then you square that and still get an integer, x must be an integer. Regardless of what that integer is.

2x= I => x = (I/2)

(I/2)^2 = I*

*a different integer

Psiva00734 · Answer

statement 1 x2 is a integer.
x can be integer or irrational numbers,
such as x=1 0r x=  2

Statement 2 2x is an integer. 
x=1or x is fractin numbers x=1/2.

both option together x=1 is a integer. 
it is not a fractin or irrational number

so option c is correct.

yazadsarkari · Answer

I forgot to test it with c

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Is x an integer?

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